Question

A proton moves perpendicularly to a uniform magnetic field B with a speed of 1.5 × 107 m/s and experiences an acceleration of 0.66 × 1013 m/s 2 in the positive x direction when its velocity is in the positive z direction d the magnitude of the field. The elemental charge is 1.60 × 10−19 C . Answer in units of T.

Answer 1

Magnetic field, B = 0.0045 T

Step-by-step explanation:

It is given that,

Speed of the proton,
`v=1.5* 10^7\ m/s`

Acceleration of the proton,
`a=0.66* 10^(13)\ m/s^2`

Charge on proton,
`q=1.6* 10^(-19)\ C`

The magnetic force is balanced by the force due to the acceleration of the proton as :

`qvB=ma`

`B=(ma)/(qv)`

`B=(1.67* 10^(-27)\ kg* 0.66* 10^(13)\ m/s^2)/(1.6* 10^(-19)\ C* 1.5* 10^7\ m/s)`

B = 0.0045 T

So, the magnitude of magnetic field on the proton is 0.0045 T. Hence, this is the required solution.